Between any two real numbers, there is an algebraic number and also a transcendental number.
I understand what algebraic numbers and transcendental numbers are, but how can I prove that they both exist between any two real numbers?

$\begingroup$For the first part just use the fact that every rational number is algebraic. Now for the second, if there are $x,y\in\mathbb R$ with $x<y$ such that $(x,y)$ contains only algebraic numbers then $(x,y)$ is countable$\endgroup$
– CIJDec 5 '16 at 1:04

Here's one way to do it without doing too much work, assuming you already know that between any two reals there is a rational:

Every rational is algebraic; so, between any to reals, there is an algebraic real.

Meanwhile, the set of algebraic numbers is countable, but every nonempty interval $(x, y)$ is uncountable. So "most" elements of $(x, y)$ are transcendental - in particular, there's at least one! So between any to reals, there's a transcendental, as well.

Here's a concrete example of how to construct a rational $q$ and a transcendental $t$ between two real numbers $r_1$ and $r_2$.

Take the decimal expansions of the two real numbers $r_1$ and $r_2$. At some point (say the $n$th digit) they must differ (otherwise they are the same number).

Now take the number composed of the first $n$ digits of $r_2$. As it has a finite decimal expansion it is rational, and clearly it falls between $r_1$ and $r_2$. Call it $q$.

To obtain a transcendental number in the range, look at the difference $r_2 - q$ and find a positive rational less than this difference (say by taking its decimal expansion as far as the first nonzero digit). Call it $d$.

$\begingroup$you mean "take a decimal expansion for each of the two real numbers..."$\endgroup$
– djechlinDec 5 '16 at 16:40

1

$\begingroup$What if the first digit of $r_2$ differing from $r_1$ is also the last non-zero digit of $r_2$? Then you would have $q = r_2$. Of course there will always be a number $r$ with finite decimal expansion such that $r_1<r<r_2$, and it should be easy enough to tweak this procedure to find such a number.$\endgroup$
– David KDec 5 '16 at 17:55